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* and then Diana Soares declared
> On Wed, 2002-02-06 at 11:22, Nick Wilson wrote:
> > 2 tables
> > content TipId | AuthId | title | keywords | desc | section | text
> > author AuthId | name | email | bio
> >
> > My question is: If I
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Hi all,
I'm planning a *very* simple DB to place tips on small biz websites on.
I'm having a little trouble conceptualizing how it might work.
My schema runs like this:
2 tables
content TipId | AuthId | title | keywords | desc | section | text
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* and then Diana Soares declared
> No. You're right!
> Unless you generate your own AuthId (with date, not auto-incremented), i
> think you will have to do a SELECT in the "author" table to get his ID
> /you will need it to insert a new tip in t
Nick,
> I don't think ...
> Am I talking rubbish?
An admission and then an invitation??? Boy you must some masochist!
You asked a question, it was answered (and there's usually more than one way to skin
the proverbial cat); and
then you appear to argue.
"Thinking" is not as good as "testing":
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* and then DL Neil declared
> > > > >
> > > > > My question is: If I enter a 'tip' and put myself as the author, when
> > > > > /I/ come to write another tip how do I make my query avoid putting
> > > > > another entry in author? I'll be using P
Nick,
> Well I /hadn't/ tested as I wasn't sure that my DB design wasn't flawed,
> I was expecting someone to say 'well if you want to do that you need to
> design like this..' No matter.
Again: if you use the native-MySQL/command line, or an administration/management
package, you will be able
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* and then [EMAIL PROTECTED] declared
> Your message cannot be posted because it appears to be either spam or
> simply off topic to our filter. To bypass the filter you must include
> one of the following words in your message:
>
> sql,query
>
Nick,
> I don't think ...
> Am I talking rubbish?
An admission and then an invitation??? Boy you must some masochist!
You asked a question, it was answered (and there's usually more than one way to skin
the proverbial cat); and
then you appear to argue.
"Thinking" is not as good as "testing":
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* and then Diana Soares declared
> No. You're right!
> Unless you generate your own AuthId (with date, not auto-incremented), i
> think you will have to do a SELECT in the "author" table to get his ID
> /you will need it to insert a new tip in t
On Wed, 2002-02-06 at 11:52, Nick Wilson wrote:
> > You can always use INSERT for the "content" table and REPLACE for the
> > "author" table.
>
> I don't think that would work as it will change the AuthId and effect
> any other 'tips' I've written.
> Each author should have one entry in 'author'
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* and then Diana Soares declared
> On Wed, 2002-02-06 at 11:22, Nick Wilson wrote:
> > 2 tables
> > content TipId | AuthId | title | keywords | desc | section | text
> > author AuthId | name | email | bio
> >
> > My question is: If I
On Wed, 2002-02-06 at 11:22, Nick Wilson wrote:
> 2 tables
> content TipId | AuthId | title | keywords | desc | section | text
> author AuthId | name | email | bio
>
> My question is: If I enter a 'tip' and put myself as the author, when
> /I/ come to write another tip how do I make my
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Hi all,
I'm planning a *very* simple DB to place tips on small biz websites on.
I'm having a little trouble conceptualizing how it might work.
My schema runs like this:
2 tables
content TipId | AuthId | title | keywords | desc | section | text
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